Problem: $\text E = \left[\begin{array}{rrr}0 & 3 & 5 \\ 5 & 5 & 2\end{array}\right]$ and $\text D = \left[\begin{array}{rr}3 & 4 \\ 3 & -2 \\ 4 & -2\end{array}\right]$ Let $\text {H = ED}$. Find $\text H$. $ {H = }$
Answer: The Strategy When multiplying matrices, we should find each entry of the resulting product matrix separately. To find entry $(i,j)$ of the resulting product matrix, we calculate the vector dot product of row $i$ of the first matrix and column $j$ of the second matrix. [I don't know what "vector dot product" is!] Finding $\text {H}_{1,1}$ $\text{H}_{1,1}$ is the dot product of the first row of $\text{E}$ and the first column of $\text{D}$. $ \text {H}=\left[\begin{array}{rr}{0} & {3} & {5} \\ 5 & 5 & 2\end{array}\right]\left[\begin{array}{rr} {3} & 4 \\ {3} & -2 \\ {4} & -2\end{array}\right]$ Therefore, this is the appropriate calculation of $\text{H}_{1,1}$. $\begin{aligned}\text{H}_{1,1}&=(0,3,5)\cdot(3,3,4)\\\\ &=0 \cdot 3+3\cdot 3 + 5\cdot 4\\\\ &=29 \end{aligned}$ The other entries of $\text{H}$ can be found similarly. Try it yourself for $\text{H}_{2,1}$ What is the appropriate calculation of ${H}_{2,1}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $5 \cdot 3+5\cdot 3 + 2\cdot 4 = 38$ (Choice B) B $3 \cdot 5 + 3\cdot 4 = 27$ (Choice C) C $0 \cdot 4+3\cdot -2 + 5\cdot -2 = -16$ Check Summary After calculating all the remaining entries of $\text{H}$, we get the following answer. $ \text {H}=\left[\begin{array}{rr}29 & -16 \\ 38 & 6\end{array}\right] $